Deriving the Formula of the Vertex of Quadratic Functions

In getting the vertex of the quadratic function in general form f(x) = ax^2 + bx + c, we usually need to convert it to the vertex form f(x) = a(x - h)^2 + k. In the latter form, the vertex of the parabola is at (h,k). For example, the function in the general form

f(x) = 2x^2 - 12x + 22

can be rewritten in the vertex form as

f(x) = 2(x - 3)^2 + 4.

In the vertex form, it is easy to see that the vertex is at (h,k) = (3,4).

Aside from this method, we can also use the ordered pair

\left ( \dfrac{-b}{2a}, \dfrac{4ac - b^2}{4a} \right )

in place of (h,k)

This means that we can get the vertex of f without changing the general form to vertex from. In this post, we are going to derive the ordered pair above. That is, we are going to show that

h = \dfrac{-b}{2a}

and

k = \dfrac{4ac - b^2}{4a}.

Expanding f(x) = a(x - h)^2 + k gives us

f(x) = a(x^2 - 2hx + h^2) + k
f(x) = ax^2 - 2ahx + ah^2 + k.

Now, the terms with both x’s in the general and vertex forms are bx and -2ahx. So, we can equate the two expressions and solve for h.

-2ahx = bx
-2ah = b
h = \frac{- b}{2a}.

Now, we can use h as the value of x in the general from. That is

f(h) = ah^2 + bh + c
f\left ( \dfrac{-b}{2a} \right ) = a \left ( \dfrac{-b}{2a} \right )^2 + b \left (\dfrac{-b}{2a} \right ) + c
= a \left (\dfrac{b^2}{4a^2} \right ) - \dfrac{-b^2}{2a} + c
= \dfrac{b^2}{4a} - \dfrac{b^2}{2a} + c
= \dfrac{b^2 - 2b^2 + 4ac}{4a}
= \dfrac{4ac - b^2}{4a}
k = \dfrac{4ac - b^2}{4a} .

This means that we can get the vertex of the function

f(x) = 2x^2 - 12x + 22

without converting it to the vertex form as follows.

h = \dfrac{-b}{2a} = \dfrac{-(-12)}{2(2)} = 3
k = \dfrac{4(2)(22) - (-12)^2}{4(2)} = \dfrac{32}{8} = 4

So, the vertex is at (h,k) = (3,4) which is consistent with our calculation above.

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